Relative frequency and combined events: a week of ready-to-teach maths

Five lessons for Year 9 Probability. Print this pack and the week is prepared: each day has a one-page plan and a student worksheet with practice and a stretch section, plus 20 weather day cards, a Venn mat, 12 event cards, a mini-check and every answer with its working.

Start here: five minutes to Monday

1. Skim the week at a glance on the next page.
2. Print the five days. Each day is two A4 sheets: a plan and a worksheet.
3. Print the two cut-out sheets, one copy per pair: sheet 1 has the 20 day cards (Days 3 to 5) and the 12 event cards (Day 5), sheet 2 the Venn mat (Days 3 and 4). Cut the cards before Day 3.
4. Each pair needs a paper cup for Day 1. Nothing else to buy.
5. Open the free interactive unit on the board. Every plan says which picture to show and which buttons to press.

The idea in plain words

When we cannot list equally likely outcomes, we estimate a probability from data. The relative frequency of an event is the number of times it happened over the number of trials, and it settles as the trials grow. With two events, sort the data into a two-way table or a Venn diagram. “A and B” is the overlap. Inclusive “A or B” is one, the other or both, and the addition rule finds it: P(A) + P(B) − P(A and B). Exclusive “A or B” is one or the other but not both: the inclusive or without the overlap.

One day, one lesson

The five lessons fill a week of maths, one lesson of about 60 minutes a day. Run them in order: each day stands on the one before. Every plan marks a split point for two shorter sessions, and every worksheet ends with a stretch section for students who finish early.

How we write these answers

A relative frequency is written as the count over the total and then as a decimal: 26/80 = 0.325. When the decimal does not end, as with a class’s own data after 30 tosses, it is rounded to three places with ≈: 7/30 ≈ 0.233. A probability estimated from data is that relative frequency. In a two-way table, the first row and column are the events happening; the four inner cells always add to the grand total.

The week at a glance

One lesson a day for a week. Each day stands on the day before, so run them in order.

How the week builds

Day 1 estimates probabilities from data; Day 2 sorts data by two events in a two-way table; Day 3 finds “and” as the overlap of a Venn diagram; Day 4 finds inclusive “or” and the addition rule; and Day 5 separates exclusive “or” from inclusive “or”. It builds on listing outcomes for compound events earlier in Year 9, and on Year 8 work with two-way tables and Venn diagrams. It prepares running simulations, which complete the Year 9 probability strand.

Materials for the week (one trip)

• From the classroom: a paper cup per pair (Day 1), scissors, envelopes for the cards, this pack printed.
• Cut-out sheet 1, one per pair: 20 day cards from a class weather record (Days 3 to 5) and 12 event cards (Day 5).
• Cut-out sheet 2, one per pair: the Venn mat the day cards are sorted onto (Days 3 and 4).
• If you have them: a plastic cup or a bottle lid works as well as a paper cup. No other equipment to buy.

Words for the word wall

trial, frequency, relative frequency, estimate, prediction, two-way table, cell, total, Venn diagram, region, overlap, neither, and, inclusive or, exclusive or, at least one, exactly one, addition rule

Dear families

This week in maths, Year 9 is estimating chances from data and combining two events with “and” and “or”. A key idea is that “or” has two meanings: one, the other or both (inclusive), or one or the other but not both (exclusive).

Try this at home

• On a walk or a drive, count how many of the next 50 cars are white. The count over 50 is a relative frequency.
• Ask: in a class of 30, 12 play netball, 8 play soccer and 3 play both. How many play neither? (13.)
• Ask: does “tea or coffee?” mean you could have both? Does “free entry if you are under 12 or over 65”? (The first usually means one of the two; in the second nobody is both, so the two kinds of or agree.)

My data week

Fill one row a day. Write something you counted, how many times it happened, out of how many, and the relative frequency.

Relative frequency

When we cannot list equally likely outcomes, we estimate a probability from data. The relative frequency of an event is the number of times it happened over the number of trials. It is an estimate, and it settles as the trials grow, so more data gives an estimate we can trust more.

We are learning to

• calculate relative frequencies from given data and from data we collect,
• use a relative frequency to estimate a probability and to predict,
• explain why more trials give a more trustworthy estimate.

Success criteria

• I can write a relative frequency as a fraction of the trials and as a decimal.
• I can explain why the class’s estimate is better than one pair’s.

You need

The worksheet; a paper cup per pair; a place on the board for each pair’s results (Part C).

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part A; start the second with the tossing.

Watch for these ideas

• Writing the count (36) instead of the relative frequency (36/50).
• Dividing by the wrong total, such as the number of other results instead of all the trials.
• Thinking 10 trials give the true probability, or that a relative frequency of 0 means impossible.

Answers

• Part A: water 26/80 = 0.325, juice 18/80 = 0.225, milk 14/80 = 0.175, hot chocolate 22/80 = 0.275; they add to 1 because every order is one of the four; P(not water) = 54/80 = 0.675; about 0.275 × 400 = 110 hot chocolates.
• Parts B and C: results vary. A pair’s relative frequency changes a lot over the first tosses; the class’s changes little as each pair is added.
• Part D: the class’s 438/600 = 0.73, from the most tosses; 0.73 × 1000 = 730 sides.

Relative frequency

A. Canteen orders

The canteen’s last 80 drink orders. Write each relative frequency as a fraction of 80 and as a decimal.

B. Toss a paper cup

Toss the cup 40 times. Tally how it lands, and after every 10 tosses work out the relative frequency of landing on its side so far.

C. Pool the class

D. Stretch: three estimates

Pair A got 9 sides in 10 tosses, pair B 27 in 40, and a whole class 438 in 600. Which is the best estimate of P(side), and why? Use it to predict the sides in 1000 tosses.

Two-way tables

A two-way table sorts data by two events at once. The four inner cells hold every case exactly once and the totals sit around the edge, so a missing count can be found from the others, and each cell over the grand total is a relative frequency.

We are learning to

• complete a two-way table from some of its counts,
• read relative frequencies from a two-way table as estimates of probabilities,
• collect data from the class into a two-way table.

Success criteria

• I can fill the gaps in a two-way table using its totals.
• I can read the probability of an event from a two-way table.

You need

The worksheet; a two-way table drawn on the board for the class survey (Part B).

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after the class survey; start the second with Part A.

Watch for these ideas

• Adding a row total into the grand total as if it were another cell.
• Reading a row total as an inner cell.
• Dividing by a row total when the question asks about everyone, not only that row.

Answers

• Part A: raining and on time 26, dry and late 30, dry and on time 120, dry total 150, on time total 146. P(late) = 54/200 = 0.27; P(raining and late) = 24/200 = 0.12; P(dry and on time) = 120/200 = 0.6.
• Part B: results vary; the four cells add to the number of students who answered.
• Part C: the gap is 1 − 0.12 − 0.18 − 0.55 = 0.15; P(cracked) = 0.12 + 0.18 = 0.3; P(has a case) = 0.12 + 0.55 = 0.67; 0.18 × 200 = 36 students.
• Part D: late on 24/50 = 0.48 of rainy trips against 30/150 = 0.2 of dry ones, so a late bus was more than twice as common when it rained.

Two-way tables

A. Late buses

A bus company checked 200 trips: was it raining, and was the bus late? Fill the gaps, then estimate each probability.

B. Our class

Two questions: “Is your first name longer than five letters?” and “Were you born from January to June?” Fill in the class’s counts, then write two relative frequencies.

C. A table of relative frequencies

A survey of Year 9 students’ phones, as relative frequencies. One cell is missing.

D. Stretch: rain and late buses

Use your table from Part A. Is a bus more likely to be late when it rains? Compare the late trips with the rainy trips, and with the dry trips.

And: the overlap

The event “A and B” means both at once. On a Venn diagram it is the overlap of the two circles; in a two-way table it is the one cell where the row and the column both say yes. Its relative frequency is that count over the total, and it can never be bigger than either event alone.

We are learning to

• sort data onto a Venn diagram and turn it into a two-way table,
• find P(A and B) from either picture,
• check that the four regions of a Venn diagram add to the total.

Success criteria

• I can place each day card in the right region of the Venn mat, including the days that are neither.
• I can find P(rain and windy) from my Venn diagram and from my table.

You need

The worksheet; per pair, the 20 day cards (cut-out sheet 1) and the Venn mat (cut-out sheet 2).

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part B; start the second with the board picture again.

Watch for these ideas

• Putting a rainy, windy day in one circle only, so the overlap is too small.
• Leaving the neither days off the mat: they go inside the box, outside both circles.
• Writing a circle’s total (28) in its only region.

Answers

• Part A: rain only 2, windy only 4, both 3, neither 11; the table has rain 5, windy 7, total 20.
• Part B: rain and windy 3/20 = 0.15, rain and calm 2/20 = 0.1, dry and windy 4/20 = 0.2, dry and calm 11/20 = 0.55; together 1, because every day is in exactly one cell.
• Part C: Japanese only 28 − 6 = 22, French only 20 − 6 = 14, neither 80 − 22 − 6 − 14 = 38; P(Japanese and French) = 6/80 = 0.075.
• Part D: no. The overlap is part of A, so it can be no bigger than A; it equals P(A) only when every A is also a B.

And: the overlap

A. Sort the weather record

Sort the 20 day cards onto the Venn mat. Write the count of each region in the Venn diagram, then fill the table.

B. Both at once

C. Two languages

Of 80 Year 9 students, 28 study Japanese, 20 study French and 6 study both. Fill the Venn diagram, then find P(Japanese and French).

D. Stretch: bigger than A?

Can P(A and B) ever be bigger than P(A)? Explain with a Venn diagram.

Inclusive or

Inclusive “or” means one, the other, or both: on a Venn diagram, everything inside either circle. Count its three regions, or use the addition rule, P(A or B) = P(A) + P(B) − P(A and B). The subtraction is there because adding the two circles counts the overlap twice.

We are learning to

• find P(A or B), inclusive, by counting the regions of the union,
• use the addition rule and explain why it subtracts the overlap,
• use the rule backwards to find the overlap.

Success criteria

• I can find P(rain or windy) two ways and get the same answer.
• I can explain why 55 + 50 is too many.

You need

The worksheet; the sorted day cards and Venn mat from Day 3 (or the Part A counts).

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part A; start the second with “The addition rule” on the board.

Watch for these ideas

• Adding P(A) and P(B) and stopping, so the overlap is counted twice.
• Taking the overlap away twice, which leaves only the two crescents: that is the exclusive or of Day 5.
• Running the rule backwards with the signs mixed up.

Answers

• Part A: counting, 2 + 3 + 4 = 9, so 9/20 = 0.45; by the rule, 5/20 + 7/20 − 3/20 = 9/20, the same.
• Part B: 28/80 + 20/80 − 6/80 = 42/80 = 0.525; 50/200 + 54/200 − 24/200 = 80/200 = 0.4; 0.45 + 0.3 − 0.1 = 0.65; 0.5 + 0.4 − 0.7 = 0.2.
• Part C: A = 4, 8, 12, 16, 20; B = 13, 14, 15, 16, 17, 18, 19, 20; A and B = 16, 20. 5/20 + 8/20 − 2/20 = 11/20 = 0.55; the list of A or B has 11 numbers too.
• Part D: 0.7 + 0.6 = 1.3 is more than 1, so at least 0.3 is counted twice: P(A and B) is at least 0.3.

Inclusive or

A. The weather record, two ways

Use your Venn mat from Day 3. Find P(rain or windy), meaning rain, wind or both, two ways.

B. The addition rule

C. Number cards

The numbers 1 to 20 are on cards, and one card is drawn. A is “a multiple of 4” and B is “more than 12”. List A, B and A and B, then find P(A or B) with the rule. Check by listing A or B.

D. Stretch: they must overlap

P(A) = 0.7 and P(B) = 0.6. Explain why A and B must overlap, and find the smallest P(A and B) can be.

Exclusive or

Exclusive “or” means one or the other but not both: the two crescents of a Venn diagram, without the overlap. It is the inclusive or minus the overlap. Everyday speech uses “or” both ways, so read the words: “or both” and “at least one” are inclusive; “but not both” and “exactly one” are exclusive.

We are learning to

• find P(A or B), exclusive, from data,
• tell and, inclusive or and exclusive or apart from the words,
• check that inclusive or minus and gives exclusive or.

Success criteria

• I can sort the twelve cards into and, inclusive or and exclusive or.
• I can find all three probabilities from one Venn diagram.

You need

The worksheet; the 12 kind cards (cut-out sheet 1) per pair; the unit’s self-check quiz.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after the sort; start the second with Part B and the quiz.

Watch for these ideas

• Reading every “or” as inclusive, or every “or” as exclusive, without looking at the words.
• Taking the overlap away once from P(A) + P(B) and calling it exclusive: that is the inclusive or.
• Missing that “neither dry nor calm” means rainy and windy.

Answers

• Part A: the card key is with the mini-check answers.
• Part B: (26 + 30)/200 = 56/200 = 0.28; check 80/200 − 24/200 = 56/200.
• Part C: dog only 64, cat only 36, both 20, neither 80; and 20/200 = 0.1, inclusive or 120/200 = 0.6, exclusive or 100/200 = 0.5.
• Part D: when the two events cannot happen together, the overlap is 0 and the two kinds of or agree; for example, one die showing a 1 or a 6.

Exclusive or

A. Which kind?

Sort the 12 cards, write each letter in its pile, and find its probability from the data below.

B. Late or raining, not both

From the 200 bus trips of Day 2 (50 rainy, 54 late, 24 both): find P(raining or late, but not both), then check it is P(raining or late) − P(raining and late).

C. All three from one diagram

Of 200 households, 84 have a dog, 56 have a cat and 20 have both. Fill the Venn diagram, then find P(dog and cat), P(dog or cat), inclusive, and P(dog or cat), exclusive.

D. Stretch: when do they agree?

When do inclusive or and exclusive or give the same probability? Give an example.

Twenty day cards and twelve kind cards

A class weather record: 20 day cards (Days 3 to 5), one set per pair

A class recorded 20 school days. Rain: it rained during the school day. Windy: at lunchtime the school flag flew straight out.

Which kind? 12 cards (Day 5), one set per pair

Sort each card: and, inclusive or, or exclusive or. Then find its probability from the data on the Day 5 worksheet.

Teacher note: the key is with the mini-check answers, so this sheet can go to pairs whole.

Venn mat for the day cards

One per pair (Days 3 and 4). Lay each day card in its region: a rainy, windy day in the overlap, and a dry, calm day inside the box but outside both circles.

What we know: Relative frequency and combined events

Work on your own. Give each relative frequency as a fraction of the total and as a decimal. Show your table or Venn diagram.

1. A bottle lid was tossed 250 times and landed top up 95 times. What is the relative frequency of top up? ______________________
2. Use your answer to predict how many times it lands top up in 1000 tosses. ______________________
3. Which is the better estimate of a probability, 7 out of 10 or 62 out of 100? Why? ______________________
4. At a carnival, 80 students took part: 44 ran a race, 30 swam a race and 14 did both. How many did neither? ______________________
5. For the carnival, find P(ran and swam). ______________________
6. Find P(ran or swam), inclusive. ______________________
7. Find P(ran or swam), exclusive. ______________________
8. P(A) = 0.6, P(B) = 0.5 and P(A or B) = 0.8. Find P(A and B). ______________________

Answers and marking guide

Answers with working

1. 95/250 = 0.38.
2. 0.38 × 1000 = 380.
3. 62 out of 100: it comes from more trials, and a relative frequency settles as the trials grow.
4. 44 + 30 − 14 = 60 did at least one, so 80 − 60 = 20.
5. 14/80 = 0.175.
6. 60/80 = 0.75.
7. (30 + 16)/80 = 46/80 = 0.575; or 60/80 − 14/80.
8. 0.6 + 0.5 − 0.8 = 0.3.

A quick three-level guide

Eight questions, four ideas. A student at standard answers most questions with a table or a Venn diagram that holds every case once.

Card key

Which kind? (Day 5): and: B (30/100 = 0.3), G (3/20 = 0.15), I (6/80 = 0.075); inclusive or: A (9/20 = 0.45), D (42/80 = 0.525), F (9/20 = 0.45), K (75/100 = 0.75), L (75/100 = 0.75); exclusive or: C (6/20 = 0.3), E (45/100 = 0.45), H (36/80 = 0.45), J (45/100 = 0.45).

Weekly class record

Jot a tick as you move around the room; the mini-check fills any gaps. A tick a day is plenty.

The five columns are the five days: relative frequency, two-way tables, and, inclusive or, and exclusive or.